dfs-Bit Compression

链接：https://www.nowcoder.com/acm/contest/145/C
来源：牛客网

Bit Compression

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld

题目描述

A binary string s of length N = 2n is given. You will perform the following operation n times :

- Choose one of the operators AND (&), OR (|) or XOR (^). Suppose the current string is S = s1s2...sk. Then, for all $1 \le i \le \frac{k}{2}$ , replace s2i-1s2i with the result obtained by applying the operator to s2i-1 and s2i. For example, if we apply XOR to {1101} we get {01}.

After n operations, the string will have length 1.

There are 3n ways to choose the n operations in total. How many of these ways will give 1 as the only character of the final string.

输入描述:

The first line of input contains a single integer n (1 ≤ n ≤ 18).The next line of input contains a single binary string s (|s| = 2n). All characters of s are either 0 or 1.

输出描述:

Output a single integer, the answer to the problem.

示例1

输入

复制

2
1001

输出

复制

说明

The sequences (XOR, OR), (XOR, AND), (OR, OR), (OR, AND) works.

题意在map那篇博客里已经说过了，这题也可用dfs做，map的做法不好想但是想出来之后代码好实现，dfs思路好想，但是代码不好实现。dfs怎么想？就是第一次可以从三个里面选一个，选完之后又可以从三个操作符里选一个，就样深搜下去。

然后自己YY了一发。

WA的代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=270000;
char s[maxn];
int FangAn;
int n;
int tmp1[maxn],tmp2[maxn],tmp3[maxn];
void dfs(int d[],int ceng,int logal,int k){int c;int cnt=0;int flag=0;if(ceng==n){if(logal==1){for(int i=1;i<=k/2;i++){d[++cnt]=(d[2*i-1]^d[2*i]);}}else if(logal==2){for(int i=1;i<=k/2;i++){d[++cnt]=(d[2*i-1] | d[2*i]);}}else{for(int i=1;i<=k/2;i++){d[++cnt]=(d[2*i-1] & d[2*i]);}}if(d[1]==1){//cout<<"FangAn"<<FangAn<<endl;FangAn++;}//cout<<d[1]<<"  %%%%  "<<endl;return ;}if(logal==1){for(int i=1;i<=k/2;i++){d[++cnt]=(d[2*i-1]^d[2*i]);cout<<cnt<<" "<<d[cnt]<<" $$ "<<endl;if(d[cnt]==1)flag=1;}c=ceng;c+=1;if(flag==0){return;}dfs(d,c,1,cnt);dfs(d,c,2,cnt);dfs(d,c,3,cnt);}else if(logal==2){for(int i=1;i<=k/2;i++){d[++cnt]=(d[2*i-1] | d[2*i]);cout<<cnt<<" "<<d[cnt]<<" YY "<<endl;if(d[cnt]==1)flag=1;}//ceng+=1;c=ceng;c+=1;cout<<c<<" "<<flag<<" TT "<<endl;if(flag==0)return;dfs(d,c,1,cnt);cout<<FangAn<<" QQ "<<endl;dfs(d,c,2,cnt);cout<<FangAn<<" WW "<<endl;dfs(d,c,3,cnt);cout<<FangAn<<" RR "<<endl;}else{for(int i=1;i<=k/2;i++){cout<<d[2*i-1]<<" "<<d[2*i]<<" ||| "<<endl;d[++cnt]=(d[2*i-1] & d[2*i]);cout<<cnt<<" "<<d[cnt]<<" +++ "<<endl;if(d[cnt]==1)flag=1;}c=ceng;c+=1;if(flag==0){return;}dfs(d,c,1,cnt);dfs(d,c,2,cnt);dfs(d,c,3,cnt);}
}
int main()
{FangAn=0;scanf("%d",&n);int length=pow(2,n);scanf("%s",s);for(int i=0;i<length;i++){if(s[i]=='0') tmp1[i+1]=0,tmp2[i+1]=0,tmp3[i+1]=0;else    tmp1[i+1]=1,tmp2[i+1]=1,tmp3[i+1]=1;}dfs(tmp1,1,1,length);cout<<FangAn<<" ### "<<endl;dfs(tmp2,1,2,length);cout<<FangAn<<" ^^^ "<<endl;//dfs(tmp3,1,3,length);cout<<FangAn<<" *** "<<endl;cout<<FangAn<<endl;}

一开始就用了一个tmp，发现运行完主函数里的dfs(tmp,1,1,length)，再输出tmp一看tmp变了，但是tmp不能变，感觉tmp不会变啊，只是让tmp去运行dfs,不知道怎么弄得，好吧既然变了我就用三个tmp来弄，这个小bug改完了。发现dfs里三个dfs摞在一起的那三个语句，执行完第一个d数组就变了，但是d数组也不能变。所以虽然是那样想，但是代码不是按想的思路走的。

所以想办法不要让d数组改变。

AC代码

#include<bits/stdc++.h>
using namespace std;
char dp[20][300000];
long long ans;
char op(int k,char a,char b){int aa=a-'0';int bb=b-'0';if(k==0) return (aa&bb)+'0';if(k==1) return (aa|bb)+'0';if(k==2) return (aa^bb)+'0';return '0';
}
void dfs(int cur){if(cur==0){if(dp[cur][0]=='1'){ans++;}return;}for(int k=0;k<3;k++){int cnt=0;for(int i=0;i<(1<<cur);i+=2){char ch=op(k,dp[cur][i],dp[cur][i+1]);if(ch=='1') cnt++;dp[cur-1][i/2]=ch;}if(cnt==0) continue;dfs(cur-1);}
}
int main(){int n;scanf("%d",&n);scanf("%s",dp[n]);ans=0;dfs(n);printf("%lld\n",ans);
}